1.4. Lecture 4

1.4.1. Schur’s Lemma

We move on to more theoretical considerations. Recall that if $G$ is a group and $(\pi,V)$ is a representation of $G$ , then $\operatorname{Hom}_{G}(V)=\operatorname{Hom}_{G}(V,V)$ is the vector space of linear maps from $V$ to $V$ that commute with every $\pi(g)$ (for all $g\in G$ ).

Lemma 1.4.1 (Schur’s Lemma, 1907).

If $(\pi,V)$ is an irreducible finite-dimensional complex representation of a group $G$ , then

\operatorname{Hom}_{G}(V)={\mathbb{C}}{\,\mathrm{Id}}.

Proof.

Let $T(\neq 0)\in\operatorname{Hom}_{G}(V)$ . Since $V$ is a finite-dimensional vector space over ${\mathbb{C}}$ , $T$ has an eigenvalue $\lambda\in{\mathbb{C}}$ . Consider the morphism $T-\lambda\,{\,\mathrm{Id}}$ . Then for any $g\in G$ ,

\pi(g)(T-\lambda{\,\mathrm{Id}})=\pi(g)T-\lambda\pi(g)=T\pi(g)-\lambda\pi(g)=(% T-\lambda{\,\mathrm{Id}})\pi(g);

$T-\lambda{\,\mathrm{Id}}$ is therefore in $\operatorname{Hom}_{G}(V)$ . By Lemma 1.2.9, $\big{(}\pi,\ker(T-\lambda{\,\mathrm{Id}})\big{)}$ is a subrepresentation of $(\pi,V)$ . The subspace $\ker(T-\lambda{\,\mathrm{Id}})\subseteq V$ is non-zero, since any $\lambda$ -eigenvector of $T$ is in $\ker(T-\lambda{\,\mathrm{Id}})$ , and so $(\pi,V)$ being irreducible then gives $V=\ker(T-\lambda{\,\mathrm{Id}})$ , hence $T=\lambda{\,\mathrm{Id}}\in{\mathbb{C}}{\,\mathrm{Id}}$ . ∎

Corollary 1.4.2.

Let $(\pi,V)$ and $(\rho,W)$ be two irreducible finite-dimensional complex representations of a group $G$ . Then

\operatorname{dim}\operatorname{Hom}_{G}(V,W)=\begin{cases}1&\text{ if $(\pi,V% )\cong(\rho,W)$}\\ 0&\text{ otherwise}.\end{cases}

Proof.

We start by assuming that the representations are not isomorphic. Given $T\in\operatorname{Hom}_{G}(V,W)$ , by Lemma 1.2.9, $(\pi,\ker(T))$ is a subrepresentation of $(\pi,V)$ and $(\rho,{\mathrm{im}}(T))$ is a subrepresentation of $(\rho,W)$ . By assumption, $(\pi,V)$ and $(\rho,W)$ are irreducible, hence

\ker(T)=0\;\mathrm{or}\;V\quad\mathrm{and}\quad{\mathrm{im}}(T)=0\;\mathrm{or}% \;W.

If ${\mathrm{im}}(T)=0$ , then $T=0$ , and we are done. If ${\mathrm{im}}(T)=W$ , then $\ker(T)\neq V$ , hence $\ker(T)=0$ . The map $T$ is therefore both injective ( $\ker(T)=0$ ) and surjective ( ${\mathrm{im}}(T)=W$ ), i.e. invertible. This contradicts the assumption that the representations are non-isomorphic; hence we must have ${\mathrm{im}}(T)=0$ .

Assume now that the representations are isomorphic. Then by assumption there exists an invertible element $T\in\operatorname{Hom}_{G}(V,W)$ . Let $S$ be any other element of $\operatorname{Hom}_{G}(V,W)$ . We then consider the map $T^{-1}S:V\rightarrow V$ . By Problem 1.2.3, $T^{-1}\in\operatorname{Hom}_{G}(W,V)$ , hence for any $g\in G$ , we have

T^{-1}S\pi(g)=T^{-1}\rho(g)S=\pi(g)T^{-1}S,

hence $T^{-1}S\in\operatorname{Hom}_{G}(V)$ . By Lemma 1.4.1, $\operatorname{Hom}_{G}(V)={\mathbb{C}}{\,\mathrm{Id}}$ , hence $T^{-1}S=\lambda{\,\mathrm{Id}}$ for some $\lambda\in{\mathbb{C}}$ . It follows that $S\in{\mathbb{C}}T$ , showing that $\operatorname{dim}\operatorname{Hom}_{G}(V,W)=1$ . ∎

1.4.2. Abelian groups

Schur’s lemma has a particularly striking application to Abelian groups:

Theorem 1.4.3.

Let $G$ be an Abelian group. Then every finite-dimensional irreducible complex representation of $G$ is 1-dimensional.

Proof.

Let $(\rho,V)$ be an irreducible representation of $G$ . For $h\in G$ , set $T_{h}=\rho(h)\in\operatorname{GL}(V)$ . Then $T_{h}\in\operatorname{Hom}_{G}(V)$ , since it commutes with $\rho(g)$ for all $g\in G$ . Indeed,

T_{h}\rho(g)=\rho(h)\rho(g)=\rho(hg)=\rho(gh)=\rho(g)\rho(h)=\rho(g)T_{h}.

Hence by Schur’s Lemma, $T_{h}=\rho(h)$ acts by a scalar $\chi(h)$ on $V$ :

\rho(h){\bf v}=\chi(h){\bf v}

for all ${\bf v}\in V$ and a non-zero scalar $\chi(h)$ . But now, any non-zero ${\bf v}\in V$ spans a $G$ -invariant subspace. Since $V$ is irreducible, this implies that $V$ is 1-dimensional, and $\chi=\rho$ is a homomorphism $G\rightarrow\mathbb{C}^{\times}$ . ∎

Remark 1.4.4.

It is possible to give an alternative proof of this using the fact from linear algebra that any commuting set of linear maps from a finite-dimensional vector space to itself has a simultaneous eigenvector.

A homomorphism $\chi:G\rightarrow\mathbb{C}^{\times}$ is often called a character (though this will later cause an unfortunate clash of notation). If $G$ is Abelian, then the group

\widehat{G}=\{\text{homomorphisms }\chi:G\rightarrow\mathbb{C}^{\times}\}

is called the character group, or dual group, of $G$ . It is a group under the operation $(\chi_{1}\chi_{2})(g)=\chi_{1}(g)\chi_{2}(g)$ .

Example 1.4.5.

Let $G=C_{n}$ be a cyclic group of order $n$ . Then $\widehat{G}\cong C_{n}$ . Indeed, pick a generator $g$ of $G$ and let $\omega=e^{2\pi i/n}$ be a primitive $n$ -th root of unity. Then a character $\chi$ of $G$ is determined uniquely by $\chi(g)$ , which must be an $n$ -th root of unity $\omega^{a}$ . If we let $\chi_{a}\in\hat{G}$ be the homomorphism such that $\chi_{a}(g)=\omega^{a}$ , then the map

a\mapsto\chi_{a}

determines a group isomorphism ${\mathbb{Z}}/n{\mathbb{Z}}\rightarrow\widehat{G}$ . This is a homomorphism because

\chi_{a+b}(g)=\omega^{a+b}=\omega^{a}\omega^{b}=\chi_{a}(g)\chi_{b}(g).

In fact, if $G$ is any finite Abelian group, then $\widehat{G}\cong G$ . You can prove this using the cyclic case and the fundamental theorem of finite Abelian groups.

For arbitrary groups $G$ the same method of proof gives:

Proposition 1.4.6.

Let $(\rho,V)$ be an irreducible finite-dimensional representation of $G$ and let

Z=Z(G)=\{z\in G\,|\,zg=gz\text{ for all $g\in G$}\}

be the center of $G$ . Then $Z$ acts on $V$ as a character: there is a character $\chi:Z\rightarrow{\mathbb{C}}^{\times}$ such that

\rho(z){\bf v}=\chi(z){\bf v}.

for all $z\in Z$ and ${\bf v}\in V$ .

We call $\chi$ the central character of $\rho$ .

Proof.

See Problem 1.4.3 below. ∎

Finally, we can use our classification of the irreducible representations of Abelian groups to get a bound on the dimension of the irreducible representations of any finite group.

Proposition 1.4.7.

Let $G$ be a finite group, $A$ be an Abelian subgroup of $G$ , and $(\pi,V)$ be an irreducible representation of $G$ . Then

\operatorname{dim}V\leq\frac{|G|}{|A|}=[G:A].

Proof.

Restricting the representation to $A$ , we find an irreducible $A$ -subrepresentation $W$ of $V$ (cf. Problem 1.4.3). By Theorem 1.4.3, $W$ is 1-dimensional, spanned by a vector ${\bf v}$ . There is then a character $\chi$ of $A$ such that

\pi(h){\bf v}=\chi(h){\bf v}

for all $h\in A$ . Now, $\{\pi(g){\bf v}\,|\,g\in G\}$ spans a non-zero subrepresentation of $V$ , hence is equal to $V$ by irreducibility. Write $g_{1}A,g_{2}A,\ldots,g_{r}A$ for the left cosets of $A$ in $G$ , where $r=[G:A]$ . Then for any $h\in A$ , we have

\pi(g_{i}h){\bf v}=\pi(g_{i})\pi(h){\bf v}=\pi(g_{i})\chi(h){\bf v}=\chi(h)(% \pi(g_{i}){\bf v})

But this implies that $V=\mathrm{span}\{\pi(g)v\,|\,g\in G\}$ is already spanned by

\{\pi(g_{i})v\,|\,1\leq i\leq r\},

so has dimension at most $r=[G:A]$ . ∎

Example 1.4.8.

The group $D_{n}$ has an abelian subgroup $C_{n}$ of index 2, and so every irreducible representation of $D_{n}$ has dimension at most $2$ .

1.4.3. Exercises

Problem 29. Let $(\pi,V)$ be a finite-dimensional representation of a group $G$ . Show that there is an irreducible subrepresentation $(\pi,W)$ of $(\pi,V)$ .

Problem 30. Let $(\pi,V)$ be an irreducible finite-dimensional representation of a group $G$ . Denoting the centre of $G$ by $Z(G)$ (i.e. $Z(G)=\{g\in G\,|\,gh=hg\quad\forall h\in G\}$ ), show that there exists a homomorphism $\psi:Z(G)\rightarrow{\mathbb{C}}^{\times}$ such that

\pi(g)=\psi(g)\,\mathrm{Id}

for all $g\in Z(G)$ . Hint: adapt the proof of Theorem 1.4.3.